1/a + 1/b + 1/c = 6/7. Find a+ b + c

[westin]

Answer is 47. I found the answer by trial and error such that a = 2, b =3, c = 42. can someone help me learn a quick solution.

my thought:

7bc + 7ac + 7ab = 6 abc

assume a <= b <= c, then 21bc >= 6 abc; meaning a<=3

if a= 1, 7bc + 7c + 7b = 6bc (since a,b,c have to be positive, this cannot be true)
if a =2, 7bc +14c + 14b = 12 bc
b + c = 5/14(bc)
from here, i know that bc has to be the multiple of 14 which means bc can be 14,28,42.....126 .... using a long trial and error method, i deduce that a =2, b =3 and c =42.

I know that there has to be a faster way to solve this. Can someone help! thanks!

 

[Steven G]

How do you know that there is a faster way?

 

[westin]

hi, this is one of the 30 questions in a 40 minute test. it took me a long time to try out all the possibilities. i am hoping someone can teach me a better way.

 

[Otis]

this is one of the 30 questions in a 40 minute test

That's only 80 seconds to read, reason and answer each question! What has your class been doing recently?

?

 

[westin]

this is the Math Count sprint round test. questions increase in difficulty and this is the last question. normally, one is really good, you may have 5 minutes left for this question. hopefully, Mr. Peterson read this and can help me think of a faster and more direct approach. =)

 

[JayJay]

Does this help https://nrich.maths.org/6541 ?

 

[westin]

thanks Jay Jay. it works. thank you. this is great help!!!

 

[westin]

6/7 = 1/2 + 5/14;
6/7 = 1/2 + 1/3 + 1/42

thank you!

 

[westin]

hi Jay Jay,

how about the following question. great, if u can give me some pointers too. thanks!!

# of ways

$ of ways Hi, the answer is 132 but I only got 126. can someone help what did I miss. thanks. # of ways from Q to U = 4 # of ways from Q to U to E = 4 x 3 =12 # of ways from Q to U to E to U = 12/2 * 4 + 12/2 * 3 = 42 ( since 4 corners E cannot be reached at this stage, only 8 E left...

www.freemathhelp.com

 

[Otis]

Does this help [Fibonacci's Greedy Algorithm]?

Nice job seeing Egyptian Fractions, JayJay.

?

 

[Dr.Peterson]

this is the Math Count sprint round test. questions increase in difficulty and this is the last question. normally, one is really good, you may have 5 minutes left for this question. hopefully, Mr. Peterson read this and can help me think of a faster and more direct approach. =)

This is a classic example of why we ask for context:

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...

www.freemathhelp.com

You presented this as if it were a problem from an algebra class (though clearly it is not just that, because of the integer requirement). What you did initially was very reasonable in such a context, particularly as it implicitly requires knowledge that the solution is unique. But a contest is different, and knowing that would have led us to look for more specialized techniques. I'm not very experienced in contest-type math, but I might have thought of Egyptian fractions if my mind had been set in that direction. (On the other hand, I wouldn't be positive that the result of the method is unique, so even though I'd have to suppose that the problem is intended to be answered with whatever answer you can find, without proving uniqueness, that grates on my mathematical sensibilities!)

Oddly, this is also an example where your (properly) showing your (algebraic) work ended up being misleading! We can't always win.

 

[westin]

Thank you Dr. Peterson. Will state the context more clearly next time and sorry about the misleading work as that's my train of thought. Learning the Egyptian fractions first time is great!

Below link is also a Math Count Sprint round context. IF you are free, please take a look and give me some pointers too.

# of ways

$ of ways Hi, the answer is 132 but I only got 126. can someone help what did I miss. thanks. # of ways from Q to U = 4 # of ways from Q to U to E = 4 x 3 =12 # of ways from Q to U to E to U = 12/2 * 4 + 12/2 * 3 = 42 ( since 4 corners E cannot be reached at this stage, only 8 E left...

www.freemathhelp.com

Thanks!!!!!

 

[Dr.Peterson]

sorry about the misleading work as that's my train of thought.

I wasn't saying you should have done anything differently; showing work is the right thing to do. I was just pointing out that sometimes that rule can have unintended consequences!

I'll try to look at the other problem later; in the mean time, try doing it again yourself, being a little more careful to identify different cases. Communicating your thinking thoroughly can help both you and others to see errors.